Question

How do you solve `\frac { 4} { v - 4} - \frac { 7} { v + 2} = \frac { 27} { v ^ { 2} - 2v - 8}`?

Answer 1

`v = 3`

Explanation:

Given:

`4/(v-4) - 7/(v+2) = 27/(v^2-2v-8)`

Note that:

`v^2-2v-8 = (v-4)(v+2)`

So, provided `v != 4` and `v != -2`, the roots of the given equation are the same as the equation given by multiplying both sides by `(v-4)(v+2)`:

`4(v+2)-7(v-4) = 27`

So:

`27 = 4(v+2)-7(v-4)`

`color(white)(27) = 4v+8-7v+28`

`color(white)(27) = -3v+36`

Add `3v-27` to both ends to get:

`3v = 9`

Divide both sides by `3` to get:

`v = 3`

Since this is neither `4` nor `-2` it is a valid solution of the original equation.

Answer 2

`color(magenta)(v=3`

Explanation:

`4/(v-4)-7/(v+2)=27/(v^2-2v-8)`

`:.4/(v-4)-7/(v+2)=27/((v+2)(v-4))`

multiply both sides by `color(magenta)(v-4`

`:.4-(7(v-4))/(v+2)=(27(cancel(v-4)))/((v+2)cancel((v-4))`

multiply both sides by `color(magenta)(v+3`

`:.4(v+2)-7(v-4)=27`

`:.4v+8-7v+28=27`

`:.-3v+36=27`

multiply both sides by `color(magenta)(-1`

`:.3v-36=-27`

`:.3v=-27+36`

`:.3v=9`

`:.v=3`

check:

substitute `color(magenta)(v=3`

`:.4/((3)-4)-7/((3)+2)=27/((3)^2-2(3)-8)`

`:.4/-1-7/5=27/(9-6-8)`

`:.-4-1.4=27/-5`

`:.color(magenta)(-5.4=-5.4`